Current time:0:00Total duration:8:05

0 energy points

# IIT JEE perpendicular plane (part 2)

Video transcript

Where we had left off, we
had used the information from these two lines to
figure out two vectors, which are really just
almost, you could just say, two position vectors
lying in this white plane. And we took their cross
product to find a vector that's perpendicular, that's normal
to this entire white plane. And that was vector d there. And vector d will have to sit,
especially to visualize it, if you start the
vector-- you can always position a vector
anywhere you want-- but if you start it at 0, it
will clearly sit in that plane. But since you could always
position it anywhere, you could say, OK,
let's start it at 0. So it will sit in
this yellow plane that we need to find
the equation for. And we know vector d sits in it. We already know that
vector a sits in it. We got that early on in the
first video on this problem. And so we could take the
cross product of d and a to find a normal vector to it. And then using
that normal vector, and then thinking about
what an arbitrary point, or what an arbitrary
vector on this plane might look like, we
can then figure out the equation for
this yellow plane. And we already touched
on it a little bit in the linear algebra
playlist, if any of this stuff makes you confused. And just a reminder, we
take the cross product of two vectors in three
dimensional space, it will give you a third vector
that is perpendicular to both. That's the whole tool that
we're using over here. So let's just take the
cross product of d and a. So this is going to be
equal to-- so for the i, we have 1 times 4, which
is 4, minus 3 times 10. So 4 minus 30 is negative 26. So we have negative 26i. And then we have
negative or minus for j. We have to swap signs for the j. And so we have negative
8 times 4 is negative 32. Negative 32 minus 2 times 10,
so minus 20, is negative 52. So we put a negative 52 here. Or we could just
say positive 52. Because it's minus
negative 52, which is the same thing as plus 52j. And finally, for the
k-- cross out that row, that column-- negative 8 times
3 is negative 24, minus 2 times 1. So minus 2, it's negative 26. So minus 26k. So this normal vector
right here or an a vector that is normal to the
yellow plane, the plane that we have to find the
equation of, is this over here. And just because it kind of
pops out at me that all of these are divisible by 26,
I'm going to define another normal vector. I don't know, let's call it--
we could call it anything. Let me call it-- I don't
know, let's call it p. Or let's call it-- I don't
know, let's call it normal 1. I mean, just to show
it's a different vector. I'm just going to divide--
although essentially it's pointing in the same direction. It has the same direction,
just different magnitude-- I'm just going to divide
all of these by negative 26. So if I divide all of
these by negative 26, I'm just scaling the vector. But it's still going to
go in the same direction. Because that's
what I care about. I care about just
finding any vector that is normal to this
yellow plane right here. I want to find any vector. I already found one. And now I'm just
going to scale it down so that it's
a simpler vector. So if I divide everything
by negative 26, I get i plus-- or I should
say, minus 2j plus k. So that might look something
like this right over there. So this is our n1 vector
that is normal to this. Now, just using this,
we can now figure out the equation of this
plane right here. So let's just think
about it a little bit. Let's just say I have a
point, or I could even say a position vector, x, y, z. And I know x, y, z
sits on this plane. So x, y, z. It sits on this plane. And we know that the point
0, 0, 0 is on the plane. So we also know that
the position vector xi plus yj plus zk have to
sit on this yellow plane in question. I'm assuming that x, y, z
is a point on this plane. I could have put
little 0's here just to make sure that--
you know, this could be a particular
point on this plane. I'm saying this is
just any point that I'm defining to be on this plane. Now, this position vector
will then be on this plane. And since this vector is normal
to this position vector right here, or another way to
say it, it's perpendicular, or it's orthogonal. If I were to take the dot
product, if I were to take that and then dot it with this
vector right over here, and dot it with that
vector right over there-- so I would take the dot product
of that and this-- yj plus zk. If I take the dot product
of these two things, it has to be equal to 0. If I take the dot product
of two vectors that are orthogonal to each
other-- and by definition, these two vectors are. I'm saying that this
guy is in the plane. This vector, we
already figured out, is perpendicular to the plane. Which means it's
perpendicular, it's orthogonal to
everything in the plane. So if we take their dot
product, it will be equal to 0. So what does that give us? So if we take their dot
product, we get x times 1. So if we take-- let me
just go term by term. So if we take x times--
the coefficient there is 1. So we have x plus
y times negative 2. So I should really
call that-- well, I could just call
that negative 2y. And then finally,
I have z times 1. There's a 1 in front of this k. It's implicitly there. So z times 1. So plus z. That's the dot product. And that has to be equal to 0. Or another way to write
it, just to simplify it a little bit, is that x minus
2y plus z is equal to 0. And that is the equation
of our plane in question. Or I should actually
write it in yellow, because it's the yellow plane. x minus 2y plus z is equal to 0. And I know this might have been
a little daunting or a little confusing, and if you actually
had to do this in exam, you wouldn't have to do
all of the explanation. But just to visualize
what we did, we used these two lines--
you can kind of view the vectors we used as kind
of the slopes of these lines in three dimensions. Don't worry about it
if that confuses you. But we used the vectors
in-- both of these specify a different vector. You could keep scaling
them into whatever. But we just took
their cross product to find a vector that is normal
to this entire white plane. And it would have to
sit in the yellow plane. Now, if it sits on
the yellow plane, we could take the cross product
of that with the first vector that we found that was
in the yellow plane. A vector that sat
on that line, or you could almost view
it as the slope of that line in
three dimensions. And if you take the
cross product of that-- it's not exactly the slope. I don't want you
all to nit-pick me. So no, it's not
exactly the slope. But a vector in
three dimensions is kind of specifying
a direction, which is analogous to a slope
in two dimensions. But if we take the cross product
of these two vectors now, we'll get a normal vector. That's what we started
this video doing. I scaled back this normal vector
just to simplify the math. And I said, OK, if I
have this normal vector, and any arbitrary vector
on this plane, when I take their dot
product it'll be 0. And so what that
allowed us to do is figure out what are the
constraints on that x, y and z. Any x, y and z that's
on the plane, when I take the dot product
of that position vector with the normal vector,
it has to be equal to 0. So that gave us the actual
constraints for the plane. Anyway, hopefully
you found that fun.